I am trying to understand the following situation
Given a real vector bundle $E\to M$ of rank r with connection $\nabla$, and its frame bundle $\pi:Fr(E)\to M$. We can define different curvatures namely:
$$F^\omega\in \Omega^2(Fr(E),Mat_r(\mathbb{R}))$$
and $$F^\nabla\in \Omega^2(M,End(E)),$$ where $\omega=\pi^*\nabla-\varphi^{-1}\circ\nabla^{can}\circ\varphi$ with $\varphi:\pi^*E\overset{\sim}{\to} Fr(E)\times \mathbb{R}^r$.
Now I want to understand how the equality: $\pi^*F^\nabla=F^\omega$ comes to stand.
I am having trouble since I am not sure how the pullback of the curvature works, whereas the pullback of the connection is clear to me. One question I thus have is when does $F^{\pi^*\nabla}=\pi^*F^\nabla$ hold/ does it even hold. if this were the case then I do not get how ${\pi^*\nabla}=\omega$ (since I thought $F^{\pi^*\nabla}=d\omega+\omega \wedge \omega$)
Any ideas?